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My text book failed to address how to find the epsilon and the delta of a Limit. Can somebody explain to me what they are and what they do?
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Hi Hurkyl,Originally posted by Hurkyl
Limits don't have an epsilon (ε) and a delta (δ) per se...
The idea of a limit is that it's supposed to capture our idea of "approaching" a number. For example, if I have the function:
f(x) = (x^{2} - 4x + 3) / (x - 1)
I know that f(1) does not exist (because 0/0 is undefined). However, if I compute some values of f near 1:
f(1.1) = -1.9
f(1.01) = -1.99
f(1.001) = -1.999
et cetera
using the epsilon (£`) and a delta (£_) is to prove the limit is logicOriginally posted by NEOclassic
Hi Hurkyl,
Also: f(.9) = -2.1
f(.99) = -2.01
f(.999) = -2.001 et cetera
and I learned in college that the validation of this technique was referred to as "Cauchy Critereon" - "For every neighborhood, no matter how small, - - -etc etc."
Of course there is no argument that the limit of the example is really - 2; and when I tried to quick fix using l'Hospital's treatment, your arduous but elegantly explicit post was verified post haste.
The Rule states that if the ratio of two functions involves division by zero at some limiting value, the ratio of the derivatives of those functions (or 2nd,3rd etc derivatives) may indicate the existance of a limit and, as I later learned, the actual value of that limit.
It's easy to see how it works for the limit (as x ->0) of [sin x]/x which is [cos x]dx/dx =1 Q.E.D.
similarly for your example the derivative ratio is [2x -4]dx/dx which when evaluated at x=1 gives [2x1 - 4] = -2 Q.E.D. Cheers, Jim
what you mean the limit does not exist??Originally posted by KL Kam
What if a limit doesn't exist? Can we or should we use the epsilon-delta argument to show a limit doesn't exist? (I think the answer is yes and we just need to proof
"For every ε>0 there exists a δ>0 such that for all x:
0 < |x - a| < δ implies |f(x) - L| <ε " isn't true)
Since we are working backward to find the values of ε
and δ, I think it is harder to do the proof, right? Would you please give me an example using epsilon-delta argument to prove a limit doesn't exist? I've never met one in class.
Thanks
Originally posted by Hurkyl
I guess a followup is in order so as not to give Prudens the impression that calculus is horribly complicated.
As NEO mentioned, there are other ways to compute limits. It is important to learn the definition of a limit so you understand what the mathematics is trying to say (and so you can get that problem right on the test), but for practical computation, different techniques are used, which you will learn in a few sections.
no...no...Originally posted by Hurkyl
You use the ε-δ definition to prove various theorems about limits and then you use those theorems to actually compute the values of limits.
For instance, for my example problem, we had
f(x) = (x^{2} - 4x + 3) / (x - 1)
However, notice the numerator can be factored into (x - 1) * (x - 3). If we let
g(x) = x - 3
then g(x) = f(x) for all x [x=] 1. We can prove, using the ε-δ definition the theorem that:
If f(x) = g(x) for all x such that p < x < a and a < x < q (for some p and q)
and f(x) -> L as x -> a
then g(x) -> L as x -> a
So then to find the limit of f(x) as x approaches 1, we simply need to find the limit of g(x) as x approaches 1.
We can also prove the following theorem, again using the ε-δ definition:
If h(x) is a polynomial, then h(x) -> h(a) as x -> a
So then we know that g(x) -> g(1) as x -> 1.
And we combine the two theorems to prove
f(x) -> -2 as x -> 1
Originally posted by newton1
no...no...
first...you already know the answer of the g(x)
second , if f(x) is unknown
how can you assume f(x)--> a and g(x)--> a
He didn't set limit value of g(x) tends to 2 as x tends to 1.you assume the f(x)--> a and g(x)--> a
because you already know the answer of f(x)
if not, you will not set the g(1)--> 2
We don't know the limit value of f(x) before step (1) and (2) but we've used e-d defination to show that we can use (1) and (2) to compute the limit value of f(x)£`-£_ definition just to prove the limit is logic when you already know the answer
L'Hospital rule is find the answer of the unknown limit
but £`-£_ definition can't do that
okOriginally posted by KL Kam
According to Hurkyl's last post, he didn't know the limit value of both f(x) and g(x) at the very beginning. He only knew f(x) can be simplified to g(x), where x [x=] 1, after factorization.
He didn't set limit value of g(x) tends to 2 as x tends to 1.
He used epsilon-delta (e-d) defination to prove the following :
If f(x) = g(x) for all x such that p < x < a and a < x < q (for some p and q) and f(x) -> L as x -> a
then g(x) -> L as x -> a ........(1)
In his example, f(x)=g(x) for all x such that p<x<1 and 1<x<q for some p and q and f(x) -> L as x ->a. (We still don't know the the limit valye, L up to now)
Then he proved, using e-d defination that
If h(x) is a polynomial, then h(x) -> h(a) as x -> a .......... (2)
By combining (1) and (2), we know that we can substitute x=1 into g(x) to get the limit value of f(x)
We don't know the limit value of f(x) before step (1) and (2) but we've used e-d defination to show that we can use (1) and (2) to compute the limit value of f(x)
I think the L'hospital rule is somehow based on the e-d defination, but not the other way round.
I'm not sure what point you're trying to make by this.ok
now i give you a question
then you try to use the £`-£_ definition to find the answer
but....don't use the L'Hospital rule to find the answer
ok??
limit [(1/x)-1/((e^x)-1)]=?? ....(x-->0)
if you can do this with £`-£_ definition
show you step
and.....don't use the other way to find the answer 1st!!!
just like what i say beforeOriginally posted by Hurkyl
I'm not sure what point you're trying to make by this.
I could just use the ε-&delta method to prove L'Hopital's rule and use L'Hopital's rule to compute the limit, and I would be fully within the constraints of your challenge because I did the whole thing using the ε-δ definition.
i am lazy to say againOriginally posted by Hurkyl
Use of L'Hopital's rule is most certainly proof that the limit has a particular value.
Hi again Hurkyl,Originally posted by Hurkyl
Use of L'Hopital's rule is most certainly proof that the limit has a particular value.
Well, maybe you can clarify specifically what you mean by those terms.i am lazy to say again
you really don't know what differeent with "prove it exist" and "find the value"
exactly!!!!!Originally posted by Integral
The fundamental definiton of a limit is in terms of an ε δ proof, but this fundamental definition and the proof which relies heavily on set theory and the basic theorms of Topology is not easily directly applied to verify the value of specfic limits. It guarentees the existance of limits if the correct condtions are met.
So, if you have an expression which satisfies the existance conditons of the proof of a limit you can be sure that the limit exists. But it does not tell you how to determine what that limit is. That is a different matter altogether. Various tools, such as L'Hoptls rule have been developed to aid in that process. The only wat to determine if a given function or series has a limit it to know how to apply the tools available and to know how to algebraiaicly manipulate the expression to perserve the limit at the same time you expose it.
This is one of those things any 5 yr onld can do..
With 10yrs experiance.